13. An introduction to AS Level organic chemistry

Locate the distinctive bond or atom pattern first, then inspect its neighbours. Connectivity distinguishes groups that contain the same atoms: –CHO, >C=O, –COOH and –COO– all contain C and O but represent different functional groups.

Check every element symbol in the formula or structure. CH₄ and CH₂=CH₂ are hydrocarbons; CH₃CH₂OH, CH₃Cl and CH₃CN are not because each contains an element other than C or H.

Ethane, CH₃CH₃, is an alkane. Ethene contains a C=C functional group, and ethanol contains an –OH functional group, so neither is an alkane even though each has a carbon chain.

The group changes electron distribution and intermolecular attractions. For example, an alcohol –OH group can form hydrogen bonds, so an alcohol and a hydrocarbon with similar size need not have similar boiling points or water solubility.

The group also identifies the part of the molecule where characteristic chemistry occurs: C=C marks an alkene, –CHO an aldehyde and –COOH a carboxylic acid. Inspect connectivity, not just which elements are present.

When converting, preserve the same atom connectivity, functional-group atoms and bond orders. In a skeletal formula, write heteroatoms and hydrogens attached to them, count each line end and vertex once, then infer only the hydrogens on carbon that are needed to complete valency four.

1. Identify the principal functional group. 2. Choose the longest parent chain that contains it and the required multiple bond. 3. Number from the end that gives the principal suffix the lowest locant; the carbon of –CHO, –COOH or –C≡N is carbon 1. 4. Add double-bond position and substituent prefixes with their locants. 5. Use commas between numbers and hyphens between numbers and words.

For an ester, name the alkyl group attached directly to the single-bonded O first; then name the chain containing the carbonyl carbon as an alkanoate. CH₃CH₂COOCH₃ is methyl propanoate, not propyl methanoate.

From a structural or displayed formula, count every written atom. From a skeletal formula, count each unlabelled line end and vertex as a carbon, keep all labelled atoms, and add the hydrogens needed to give each carbon four bonds. Multiple bonds use two or three of those bonds. Finally, divide every molecular-formula subscript by their greatest common divisor only if an empirical formula is requested.

A nucleophile donates an electron pair to form a bond; it has an available lone pair or electron-rich bond. An electrophile accepts an electron pair; it is positively charged or electron-deficient. Nucleophilic and electrophilic describe a species' role in a reaction.

For nucleophilic substitution of a halogenoalkane, draw one arrow from the nucleophile's lone pair to the carbon bonded to X and a second arrow from the C–X bond to X. The two arrows account for bond formation and bond breaking.

For nucleophilic addition to C=O, draw an arrow from the nucleophile's lone pair to the carbonyl carbon and another from the C=O π bond to oxygen. For electrophilic addition, the first arrow instead begins at the C=C π bond and points to the electrophile.

Hybrid orbitals point as far apart as possible, which minimises repulsion between bonding electron regions. Two directions lie opposite, three spread in one plane, and four point towards the corners of a tetrahedron.

Assign the geometry around the atom being considered: a carbon in C≡C is sp, a carbon in C=C is sp², and a carbon with four single-bond directions is sp³. The label describes a local bonding environment, not automatically the shape of the whole molecule.

Every bonded pair of atoms has one σ bond: a single bond is 1σ, a double bond is 1σ + 1π, and a triple bond is 1σ + 2π. Ethene therefore has five σ bonds and one π bond; ethyne has three σ bonds and two π bonds.

The three σ-bond directions around each sp² carbon lie in one plane. The remaining p orbitals stand perpendicular to that plane and must stay parallel for sideways overlap, so rotation about C=C would destroy the π overlap.

State which atoms are planar rather than calling an entire large molecule flat. A molecule may contain a planar alkene region alongside sp³ atoms whose bonds point out of that plane.

First confirm identical molecular formulae, then compare connectivity and classify the first structural feature that differs. Different names alone do not prove the isomer type.

Always compare connectivity first. If it differs, the pair is structural rather than stereo; only then test for restricted rotation or a chiral centre. E/Z nomenclature is acceptable but not required here.

Each alkene carbon must be attached to two different groups. When a corresponding pair lies on the same side the isomer is cis; when it lies on opposite sides it is trans. If cis/trans labels are ambiguous, E/Z may be used but is not required.

A compound can contain more than one chiral centre. Check each tetrahedral carbon independently by tracing all four attached groups until their first point of difference; this section does not require meso or diastereoisomer nomenclature.

For every candidate, recount every element, complete four bonds around carbon and compare connectivity with structures already listed. Only after unique structural isomers are fixed should separate stereoisomers be counted.